Azumaya algebras of a ring with a finite automorphism group.

*(English)*Zbl 0845.16030Let \(G\) be a finite group of automorphisms of a ring \(R\), let \(R^G\) be the subring of invariant elements of \(R\), let \(C\) be the center of \(R\) and \(C^G = C \cap R^G\), and let \(V\) be the centralizer (i.e. commutant) of \(R^G\) in \(R\). The main results of this paper may be summarized as follows. If \(R\) is a separable algebra over \(C\) and \(R^G\) is a central, separable algebra over \(C^G\); then \(R\) is the tensor product over \(C^G\) of its subrings \(R^G\) and \(V\), and \(V\) is a central, separable algebra over \(C\). Now assume that \(R\) is the tensor product over \(C^G\) of its subrings \(R^G\) and \(V\). Then \(R\) is a separable algebra over \(C\) if, and only if, \(R^G \cdot C\) and \(V\) are central, separable algebras over \(C\). The authors investigate some conditions which imply that \(R\) is the tensor product of its subrings \(R^G\) and \(V\).

In theorem 3.4 of the paper, it seems that the hypothesis that some element of \(C\) has trace 1 may be replaced by an assumption that some element of \(V\) has trace 1; and this modification may be needed for the proof of theorem 3.5. Perhaps because of misprints, example 2 of the paper makes little sense.

In theorem 3.4 of the paper, it seems that the hypothesis that some element of \(C\) has trace 1 may be replaced by an assumption that some element of \(V\) has trace 1; and this modification may be needed for the proof of theorem 3.5. Perhaps because of misprints, example 2 of the paper makes little sense.

Reviewer: H.F.Kreimer (Tallahassee)

##### MSC:

16W20 | Automorphisms and endomorphisms |

16H05 | Separable algebras (e.g., quaternion algebras, Azumaya algebras, etc.) |